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Home » From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre? (1) (2) (3) (4)
From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre? (1) 15 MR ^{2} / 32 (2) 13 MR ^{2} / 32 (3) 11 MR ^{2} / 32 (4) 9 MR ^{2} / 32
Physics
From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre? (1) 15 MR ^{2} / 32 (2) 13 MR ^{2} / 32 (3) 11 MR ^{2} / 32 (4) 9 MR ^{2} / 32

Answer: Option 2)

Solving using the concept of cut part: \mathrm{I}_{\text {remain }}=\mathrm{I}-\mathrm{I}_{\mathrm{cat}}

I_{cut}=\frac{M}{\pi R^{2}} \times \frac{\pi R^{2}}{4} \times\left(\frac{R}{2}\right)^{2}{2}+\frac{M}{4} \times\left(\frac{R}{2}\right)^{2}

I=\frac{M R^{2}}{2}

    \[\begin{aligned} &=\frac{\mathrm{MR}^{2}}{2}-\left[\frac{\frac{\mathrm{M}}{\pi \mathrm{R}^{2}} \times \frac{\pi \mathrm{R}^{2}}{4} \times\left(\frac{\mathrm{R}}{2}\right)^{2}}{2}+\frac{\mathrm{M}}{4} \times\left(\frac{\mathrm{R}}{2}\right)^{2}\right] \\ &=\frac{\mathrm{MR}^{2}}{2}-\left[\frac{\mathrm{M}}{32}+\frac{\mathrm{M}}{16}\right] \\ &=\mathrm{MR}^{2}\left(\frac{1}{2}-\frac{3}{32}\right)=\frac{13 \mathrm{MR}^{2}}{32} \end{aligned}\]

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